how i can draw a sector of a circle in matlab?
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vatankhah
le 15 Déc 2014
Réponse apportée : Hamidullah Riaz
le 19 Août 2021
I want to draw a sector of a circle. The angle and the coordinate of sector's center are specified but the direction of the angle is random. i should mention that i need this pie to be a complete one, not just the curved part. like one of the pieces in this command:
pie([2 4 3 5],{'North','South','East','West'})
could anyone help me for the code please?
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Roger Stafford
le 16 Déc 2014
Let r be the circle's radius, P0 = [x0,y0] be its center, and theta be the required angle in radians.
a1 = 2*pi*rand; % A random direction
a2 = a1 + theta;
t = linspace(a1,a2);
x = x0 + r*cos(t);
y = y0 + r*sin(t);
plot([x0,x,x0],[y0,y,y0],'y-')
axis equal
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Roger Stafford
le 25 Mai 2016
@SaN AruL: Look at the description of the problem, “Let r be the circle's radius, P0 = [x0,y0] be its center, and theta be the required angle in radians.” Theta, along with the radius, r, and the center P0 = [x0,y0] are assumed to be specified prior to the execution of the code. Theta is the assumed angle the circular sector.
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Adam
le 15 Déc 2014
I don't have time to give complete code and it isn't an area I have much expertise in, but the following example shows how you can plot a sector.
t = linspace(0,0.5*pi,128);
x = [0 cos(t) 0];
y = [0 sin(t) 0];
figure; patch( x, y, 'r' )
You can obviously change the range in t to suit start and end angles as you please and use the help to look further into patch or this blog post may provide some further help even though it heads off in a different direction:
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Hamidullah Riaz
le 19 Août 2021
With a little manipulation:
x0=0;
y0=0;
theta=0.25*pi;
a1 = 2*pi*rand; % A random direction
r=100; % radius
a2 = a1 + theta;
t = linspace(a1,a2);
x = x0 + r*cos(t);
y = y0 + r*sin(t);
plot([x0,x,x0],[y0,y,y0],'y-')
axis equal
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